專業(yè)英語翻譯實訓(xùn)計劃書.doc

2012/2013學(xué)年上期專業(yè)英語翻譯實訓(xùn) 計劃書 姓名： 班級： 學(xué)號： 時 間班 級地 點指導(dǎo)教師一、實習(xí)（訓(xùn)）目的及要求1通過實訓(xùn)，使學(xué)生能借助詞典等工具，初步具備閱讀與本專業(yè)相關(guān)的英文文章的能力；熟悉科技英語翻譯的一般技巧和基本技能；掌握閱讀和翻譯專業(yè)科技資料的能力和方法。2翻譯材料由老師指定。要求詞匯量不得少于3000單詞；翻譯內(nèi)容原則上須選用計算機控制技術(shù)、DCS系統(tǒng)、儀器儀表和自動控制原理及系統(tǒng)、火電廠熱能動力設(shè)備及系統(tǒng)等與本專業(yè)相關(guān)的英語教材、教參、科技報告、專業(yè)期刊、產(chǎn)品說明書等。3翻譯要求尊重原文，用詞準確，語句通順，邏輯清楚。4為便于規(guī)范和管理，實訓(xùn)報告建議打印（同時交電子稿）。格式要求：標題格式為標題2；正文字號為小四，字體為仿宋。英文字體為Times New Roman，字號為小四。二、實習(xí)（訓(xùn)）內(nèi)容翻譯本專業(yè)相關(guān)的英語資料。三、實習(xí)（訓(xùn)）進程表1.熱控10班：時 間實訓(xùn)任務(wù)實訓(xùn)地點12.24實訓(xùn)動員，并安排實訓(xùn)任務(wù) 教室12.2512.27翻譯英語資料、編寫和打印實訓(xùn)報告教室，圖書館電子閱覽室12.28提交實習(xí)報告；答辯教室，圖書館電子閱覽室四、紀律要求實習(xí)期間須嚴格遵守校規(guī)校紀，按要求作息。不得無故遲到、早退、曠課。凡缺課時間超過實訓(xùn)總學(xué)時三分之一以上，取消實訓(xùn)資格。五、成績評定方法1考勤占20；2實習(xí)報告和答辯占80。Section 2.8 Design ExamplesTherefore, the output measurement isA plot of yt) is shown in Figure 2.46 for P=3. We can see that y(t) is proportional to the force after 5 seconds. Thus in steady state, after 5 seconds, the response y i) is proportional to the acceleration, as desired. If this period is excessively long, we must increase the spring constant, k, and the friction, b, while reducing the mass, M. If we are able to select the components so that b/M=12 and k/M =32, the accelerometer will attain the proportional response in 1 second. (It is left to the reader to show this.) EXAMPLE 2.16 Design of a laboratory robot In this example, we endeavor to show the physical design of a laboratory device and demonstrate its complex design. We will also exhibit the many components commonly used in a control system. A robot for laboratory use is shown in Figure 2.47. A laboratory robots work volume must allow the robot to reach the entire bench area and access existing analytical instruments. There must also be sufficient area for a stockroom of supplies for unattended operation.The laboratory robot can be involved in three types of tasks during an analytical experiment. The first is sample introduction, wherein the robot is trained to accept a number of different sample trays, racks, and containers and to introduce them into the system. The second set of tasks involves the robot transporting the samples between individual dedicated automated stations for chemical preparation and instrumental analysis. Samples must be scheduled and moved between these stations as necessary to complete the analysis. In the third set of tasks for the robot, flexible automation provides new capability to the analytical laboratory. The robot must be programmed to emulate the human operator or work with various devices. All of these types of operations are required for an effective laboratory robot. The ORCA laboratory robot is an anthropomorphic arm, mounted on a rail, designed as the optimum configuration for the analytical laboratory 14. The rail can be located at the front or back of a workbench, or placed in the middle of a table when access to both sides of the rail is required. Simple software commands permit moving the arm from one side of the rail to the other while maintaining the wrist position(to transfer open containers) or locking the wrist angle (to transfer objects in virtually any orientation). The rectilinear geometry, in contrast to the cylindrical geometry used by many robots, permits more accessories to be placed within the robot workspace and provides an excellent match to the laboratory bench. Movement of all joints is coordinated through software, which simplifies the use of the robot by representing the robot positions and movements in the more familiar Cartesian coordinate space.Chapter 2 Mathematical Models of SystemsFIGURE 2.47 Laboratory robot used for sample preparation. The robot manipulates small objects, such as test tubes, and probes in and out of tight places at relatively high speeds 15. ( Photo courtesy of Beckman Coulter, Inc. ) Table 2.9 ORCA Robot Arm Hardware SpecificationsArticulated, Joy Stick withArm Rail-Mounted Teach Pendant Emergency Stop Degrees of freedom 6 Cycle lime 4 s (move 1 inch up, 12 inch across, 1 inch down, and back) Reach 54 cm Maximum speed 75 cm/sHeight 78 cm Dwell time 50 ms typical (for moves within a motion)Rail 1 and 2 m Payload 0.5 kg continuous, 2.5 kg transient (with restrictions)Weight 8.0 kg Vertical deflection 1.5 mm at continuous payload 1Precision Cross-sectional 0.25 mm work envelopeFinger travel 40 mm(gripper)Gripper rotation 77 revolutions The physical and performance specifications of the ORCA system are shown in Table 2.9. The design for the ORCA laboratory robot progressed to the selection of the component parts required to obtain the total system. The exploded view of the component parts required to obtain the total system. The exploded view of the robot is shown in Figure 2.48. This device uses six DC motors, gears, belt drives, and a rail and carriage. The specifications are challenging and require the designer tomodel the system components and their interconnections accurately.Section 2.8 Design ExamplesFIGURE 2.48 Exploded view of the ORCA robot showing the components 15. ( Courtesy of Beckman Coulter, Inc.)I1 = (Vi - Vi) G,I1= ( V2 - V3) G,V2= (I1- I2) R,V3 = I2Z,Chapter 2 Mathematical Models of SystemsFIGURE 2.49 (a) Ladder network, (b) its signal-flow graph, and (c) its block diagram.where G = 1/R, Z(s) = l/Cs, and I1 (s) = I1 (we omit the (s). The signal-flow graph constructed for the four equations is shown in Figure 2.49(b), and the corresponding block diagram is shown in Figure 2.49(c). The three loops are L1= -GR = -1, L2 = -GR =-1,and L3 = -GZ. All loops touch the forward path. Loops Li and L3 are nontouching. Therefore, the transfer function isIf one prefers to utilize block diagram reduction techniques, one can start at the output withV3(s) = ZI2(s).But the block diagram shows thatI2(s) = G (V2(s) - V3(s).Therefore,V2(s) = ZGV2(s) - ZGV3(s)Section 2.9 The Simulation of Systems Using Control Design SoftwareSoWe will use this relationship between V3(s) and Vz(s) in the subsequent development. Continuing with the block diagram reduction, we havebut from the block diagram, we see thatTherefore, Substituting for V2(s) yieldsBut we know that GR = 1; hence, we obtainNote that the DC gain is 1/2, as expected. The pole is desired at p = 2(106.1) = 666.7 = 2000/3. Therefore, we require RC = 0.001. Select R = 1k and C = 1F. Hence, we achieve the filter2.9 THE SIMULATION OF SYSTEMS USING CONTROL DESIGN SOFTWAREApplication of the many classical and modern control system design and analysis ools is based on mathematical models. Most popular control design software packages an be used with systems given in the form of transfer function descriptions. In his book, we will focus on m-file scripts containing commands and functions to analyze and sign control systems. Various commercial control system packages re available for student use. The m-files described here are compatible with the ATLAB ontrol System Toolbox and the LabVIEW MathScript RT Module.See Appendix A for an introduction to MATLAB.See Appendix B for an introduction to LabVIEW MathScipt RT Module.Chapter 2 Mathematical Models of SystemsWe begin this section by analyzing a typical spring-mass-damper mathematical model of a mechanical system. Using an m-file script, we will develop an interactive analysis capability to analyze the effects of natural frequency and damping on the unforced response of the mass displacement. This analysis will use the fact that we haVe an analytic solution that describes the unforced time response of the mass displacement.Later, we will discuss transfer functions and block diagrams. In particular, we are interested in manipulating polynomials, computing poles and zeros of transfer functions, computing closed-loop transfer functions, computing block diagram reductions, and computing the response of a system to a unit step input. The section concludes with the electric traction motor control design of Example 2.14.The functions covered in this section are roots, poly, conv, polyval, tf, pzmap, pole, zero, series, parallel, feedback, minreal, and step. Spring-Mass-Damper System. A spring-mass-damper mechanical system is shown in Figure 2.2. The motion of the mass, denoted by y(t), is described by the differential equation The unforced dynamic response yt) of the spring-mass-damper mechanical system is where and The initial displacement is y(0). The transient system response is underdamped when 1, and critically damped when= 1. We can visualize the unforced time response of the mass displacement following an initial displacement of y(0). Consider the underdamped case: The commands to generate the plot of the unforced response are shown in Figure 2.50. In the setup, the variables y(0)，t ,and are input at the command level. Then the script unforced.m is executed to generate the desired plots. This creates an interactive analysis capability to analyze the effects of natural frequency and damping on the unforced response of the mass displacement. One can investigate the effects of the natural frequency and the damping on the time response by simply entering new values ofand at the command prompt and running the script unforced.m again. The time-response plot is shown in Figure 2.51. Notice that the script automatically labels the plot with the values of the damping coefficient and natural frequency. This avoids confusion when making many interactive simulations. Using scripts is an important aspect of developing an effective interactive design and analysis capability For the spring-mass-damper problem, the unforced solution to the differential equation was readily available. In general, when simulating closed-loop feedbackSection 2.9 The Simulation of Systems Using Control Design Softwarey0=0.15;nwn=sqrt(2); zeta=1/(2*sqrt(2);t=0:0.1:10;unforcedunforced.mFIGURE 2.50 Script to analyze the spring-massdamperFIGURE 2.51 Spring-massdamper Unforced response.control systems subject to a variety of inputs and initial conditions, it is difficult to obtain the solution analytically. In these cases, we can compute the solutions numerically and to display the solution graphically.Most systems considered in this book can be described by transfer functions. Since the transfer function is a ratio of polynomials, we begin by investigating how to manipulate polynomials, remembering that working with transfer functions means that both a numerator polynomial and a denominator polynomial must be specified Chapter 2 Mathematical Models of Systems FIGURE 2.52 Entering the Polynomial p(p) = s3 + 3s2 + 4 and calculating its roots.Polynomials are represented by row vectors containing the polynomial coefficients in order of descending degree. For example, the polynomialis entered as shown in Figure 2.52. Notice that even though the coefficient of the sterm is zero, it is included in the input definition of p(s)If p is a row vector containing the coefficients of p(s) in descending degree, then roots(p) is a column vector containing the roots of the polynomial. Conversely, if r is a column vector containing the roots of the polynomial, then poly(r) is a row vector with the polynomial coefficients in descending degree. We can compute the roots of the polynomial p(s) = s3, + 3s2 + 4 with the roots function as shown in Figure 2.52. In this figure, we show how to reassemble the polynomial with the poly function.Multiplication of polynomials is accomplished with the conv function. Suppose we want to expand the polynomialThe associated commands using the conv function are shown in Figure 2.53. Thus, the expanded polynomial isFIGURE 2.53 Using conv and polyval to multiply and evaluate the polynomials (3s + 2s + 1) (s + 4). Section 2.9 The Simulation of Systems Using Control Design SoftwareFIGURE 2.54 (a) The tf function. (b) Using the tf function to create transfer function objects and adding them using t h e + operator.The function polyval is used to evaluate the value of a polynomial at the given value of the variable. The polynomial n(s) has the value n(5) = -66, as shown in Figure 2.53.Linear, time-invariant system models can be treated as objects, allowing one to manipulate the system models as single entities. In the case of transfer functions, one creates the system models using the tf function; for state variable models one employs The ss function (see Chapter 3). The use of tf is illustrated in Figure 2.54(a). For example, if one has the two system modelsone can add them using the + operator to obtainThe corresponding commands are shown in Figure 2.54(b) where sysl represents Gi(s) and sys2 represents G2Cs). Computing the poles and zeros associated with a transfer function is accomplished by operating on the system model object with the pole and zero functions, respectively, as illustrated in Figure 2.55.In the next example, we will obtain a plot of the pole-zero locations in the complex plane. This will be accomplished using the pzmap function, shown in Figure 2.56. On the pole-zero map, zeros are denoted by an o and poles are denoted by an X .If the pzmap function is invoked without left-hand arguments, the plot is generated automatically.Chapter 2 Mathematical Models of SystemsFIGURE 2.55 (a) The pole and zero functions. (b) Using the pole and zero functions to compute the pole and zero locations of a linear system.FIGURE 2.56 The pzmap function.EXAMPLE 2.18 Transfer functionsConsider the transfer functionsUsing an m-file script, we can compute the poles and zeros of G(s), the characteristic equation of H(s), and divide G(s) by H(s). We can also obtain a plot of the pole-zero map of G(s)IH(s) in the complex plane.The pole-zero map of the transfer function G(s)IH(s) is shown in Figure 2.57, and the associated commands are shown in Figure 2.58. The pole-zero map shows clearly thefive zero locations, but it appears that there are only two poles. This Section 2.9 The Simulation of Systems Using Control Design SoftwareFIGURE 2.57 Pole-zero map for G(s)/H(s).FIGURE 2.58 Transfer function example for Gs) and H(s).cannot be the case, since we know that for physical systems the number of poles must be greater than or equal to the number of zeros. Using the roots function, we can ascertain that there are in fact four poles at s = 1. Hence, multiple poles or multiple zeros at the same location cannot be discerned on the pole-zero map.Chapter 2 Mathematical Models of SystemsFIGURE 2.59 Open-loop control system (without feedback).Block Diagram Models. Suppose we have developed mathematical models in the form of transfer functions for a process, represented by G(s), and a controller, represented by Gc(s), and possibly many other system components such as sensors and actuators. Our objective is to interconnect these components to form a control system.A simple open-loop control system can be obtained by interconnecting a process and a controller in series as illustrated in Figure 2.59. We can compute the transfer function from R(s) to Y(s), as follows.EXAMPLE 2.19 Series connectionLet the process represented by the transfer function G(s) beand let the controller represented by the transfer function Gc(s) beWe can use the series function to cascade two transfer functions G(s) and G2(s), as shown in Figure 2.60.The transfer function Gc(s)G(s) is computed using the series function as shown in Figure 2.61 .The resulting transfer function iswhere sys is the transfer function name in the m-file script.FIGURE 2.60 (a) Block diagram. (b) The series function.Section 2.9 The Simulation of Systems Using Control Design SoftwareFIGURE 2.61 Application of the series function.FIGURE 2.62 (a) Block diagram. (b) The parallel function.FIGURE 2.63 A basic control system with unity feedback.Block diagrams quite often have transfer functions in parallel. In such cases, the function parallel can be quite useful. The parallel function is described in Figure 2.62.We can introduce a feedback signal into the control system by closing the loop with unity feedback, as shown in Figure 2.63. The signal Ea(s) is an error signal; the signal R(s) is a reference input. In this control system, the controller is in the forward path, and the closed-loop transfer function isChapter 2 Mathematical Models of SystemsFIGURE 2.64 (a) Block diagram. (b) The feedback function with unity feedback.FIGURE 2.65 (a) Block diagram. (b) The feedback function.We can utilize the feedback function to aid in the block diagram reduction process to compute closed-loop transfer functions for single- and multiple-loop control systems.It is often the case that the closed-loop control system has unity feedback, as illustrated in Figure 2.63. We can use the feedback function to compute the closedloop transfer function by setting H(s) = 1. The use of the feedback function for unity feedback is depicted in Figure 2.64.The feedback function is shown in Figure 2.65 with the associated system configuration, which includes H(s) in the feedback path. If the input sign is omitted, then negative feedback is assumed.Section 2.9 The Simu